Fundamental theorem of calculus

From Wikipedia, the free encyclopedia

Topics in calculus
Fundamental theorem Limits of functions Continuity Vector calculus Tensor calculus Mean value theorem
Differentiation
Product rule Quotient rule Chain rule Implicit differentiation Taylor's theorem Related rates Table of derivatives
Integration
Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution

The fundamental theorem of calculus is the statement that the two central operations of calculus, differentiation and integration, are inverse operations: if a continuous function is first integrated and then differentiated, the original function is retrieved. An important consequence, sometimes called the second fundamental theorem of calculus, allows one to compute integrals by using an antiderivative of the function to be integrated. The first published statement and proof of the fundamental theorem was by the Scottish mathematician James Gregory (1638-1675).^[1]

1 Intuition
2 Formal statements
3 Corollary
4 Examples
5 Proof
- 5.1 Alternative proof
6 Generalizations
7 See also
8 Notes
9 References

Intuitively, the theorem simply states that the sum of infinitesimal changes in a quantity over time (or some other quantity) add up to the net change in the quantity.

To comprehend this statement, we will start with an example. Suppose a particle travels in a straight line with its position given by x(t) where t is time. The derivative of this function is equal to the infinitesimal change in quantity per infinitesimal change in time (of course, the derivative itself is dependent on time). Let us define this change in distance per time as the speed v of the particle. In Leibniz's notation:

$\frac{\mathrm dx}{\mathrm dt} = v(t).$

Rearranging this equation (see differential (infinitesimal)), it follows that:

$\mathrm dx = v(t)\,\mathrm dt.$

By the logic above, a change in x, call it $Δ x$ , is the sum of the infinitesimal changes dx. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; hence, the integration operation allows the recovery of the original function from its derivative. As one can reasonably infer, this operation works in reverse as we can differentiate the result of our integral to recover the original derivative.

The theorem is perhaps the most well known Fundamental theorem in popular culture. To impress the judges in a contest, Donald Duck figure skates the theorem on ice. The theorem is written differently and is slightly incorrect: one of the equal signs should have been a minus sign. Donald used f'(x) instead of f'(x)dx, which is usually frowned upon, but not improper. Nothing in the notation indicates the vector analysis version either. The correct version for real functions should have been : $\int_b^a f'(x) \mathrm d x = f(a)-f(b)$

Stated formally, the first (part of the) fundamental theorem of calculus says the following.

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined for x in [a, b] by

$F(x) = \int_a^x f(t)\, \mathrm dt$

then

$F'(x) = f(x)\,$

for every x in [a, b].

The second (part of the) fundamental theorem of calculus is a kind of converse to this.

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be a function such that

$f(x) = F'(x)\,$ for all x in [a, b]

then

$\int_a^b f(x)\,\mathrm dx = F(b) - F(a)$ .

Let f be a real-valued function defined on a closed interval [a, b]. Let F be a function such that

$f(x) = F'(x)\,$ for all x in [a, b]

then

$F(x) = \int_a^x f(t)\,\mathrm dt + F(a)$ for all x in [a, b]

and

$f(x) = \frac{\mathrm d}{\mathrm dx} \int_a^x f(t)\,\mathrm dt$ .

As an example, suppose you need to calculate

$\int_2^5 x^2\, \mathrm dx.$

Here, $f (x) = x 2$ and we can use $F(x) = {x^3\over 3}$ as the antiderivative. Therefore:

$\int_2^5 x^2\, \mathrm dx = F(5) - F(2) = {125 \over 3} - {8 \over 3} = {117 \over 3} = 39.$

Or, more generally, suppose you need to calculate

${\mathrm d \over \mathrm dx} \int_0^x t^3\, \mathrm dt.$

Here, $f (t) = t 3$ and we can use $F(t) = {t^4 \over 4}$ as the antiderivative. Therefore:

${\mathrm d \over \mathrm dx} \int_0^x t^3\, \mathrm dt = {\mathrm d \over \mathrm dx} F(x) - {\mathrm d \over \mathrm dx} F(0) = {\mathrm d \over \mathrm dx} {x^4 \over 4} = x^3.$

But this result could have been found much more easily as

${\mathrm d \over \mathrm dx} \int_0^x t^3\, \mathrm dt = f(x) {\mathrm dx \over \mathrm dx} - f(0) {\mathrm d0 \over \mathrm dx} = x^3.$

Suppose that

$F(x) = \int_{a}^{x} f(t) \mathrm dt.$

Let there be two numbers x₁ and x₁ + Δx in [a, b]. So we have

$F(x_1) = \int_{a}^{x_1} f(t) \mathrm dt$

and

$F(x_1 + \Delta x) = \int_{a}^{x_1 + \Delta x} f(t) \mathrm dt.$

Subtracting the two equations gives

$F(x_1 + \Delta x) - F(x_1) = \int_{a}^{x_1 + \Delta x} f(t) \mathrm dt - \int_{a}^{x_1} f(t) \mathrm dt. \qquad (1)$

It can be shown that

$\int_{a}^{x_1} f(t) \mathrm dt + \int_{x_1}^{x_1 + \Delta x} f(t) \mathrm dt = \int_{a}^{x_1 + \Delta x} f(t) \mathrm dt.$

(The sum of the areas of two adjacent regions is equal to the area of both regions combined.)

Manipulating this equation gives

$\int_{a}^{x_1 + \Delta x} f(t) \mathrm dt - \int_{a}^{x_1} f(t) \mathrm dt = \int_{x_1}^{x_1 + \Delta x} f(t) \mathrm dt.$

Substituting the above into (1) results in

$F(x_1 + \Delta x) - F(x_1) = \int_{x_1}^{x_1 + \Delta x} f(t)\mathrm dt. \qquad (2)$

According to the mean value theorem for integration, there exists a c in [x₁, x₁ + Δx] such that

$\int_{x_1}^{x_1 + \Delta x} f(t) \mathrm dt = f(c) \Delta x$ .

Substituting the above into (2) we get

$F(x_1 + \Delta x) - F(x_1) = f(c) \Delta x \,$ .

Dividing both sides by Δx gives

$\frac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = f(c).$

Notice that the expression on the left side of the equation is Newton's difference quotient for F at x₁.

Take the limit as Δx → 0 on both sides of the equation.

$\lim_{\Delta x \to 0} \frac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = \lim_{\Delta x \to 0} f(c).$

The expression on the left side of the equation is the definition of the derivative of F at x₁.

$F'(x_1) = \lim_{\Delta x \to 0} f(c). \qquad (3)$

To find the other limit, we will use the squeeze theorem. The number c is in the interval [x₁, x₁ + Δx], so x₁ ≤ c ≤ x₁ + Δx.

Also, $\lim_{\Delta x \to 0} x_1 = x_1$ and $\lim_{\Delta x \to 0} x_1 + \Delta x = x_1$ .

Therefore, according to the squeeze theorem,

$\lim_{\Delta x \to 0} c = x_1.$

Substituting into (3), we get

$F'(x_1) = \lim_{c \to x_1} f(c).$

The function f is continuous at c, so the limit can be taken inside the function. Therefore, we get

$F'(x_1) = f(x_1) \,$ .

which completes the proof.

(Leithold et al, 1996)

This is a limit proof by Riemann sums.

Let f be continuous on the interval [a, b], and let F be an antiderivative of f. Begin with the quantity

$F(b) - F(a)\,$ .

Let there be numbers

x₁, ..., x_n

such that

$a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b$ .

It follows that

$F(b) - F(a) = F(x_n) - F(x_0) \,$ .

Now, we add each F(x_i) along with its additive inverse, so that the resulting quantity is equal:

$\begin{matrix} F(b) - F(a) & = & F(x_n)\,+\,[-F(x_{n-1})\,+\,F(x_{n-1})]\,+\,\ldots\,+\,[-F(x_1) + F(x_1)]\,-\,F(x_0) \, \\ & = & [F(x_n)\,-\,F(x_{n-1})]\,+\,[F(x_{n-1})\,+\,\ldots\,-\,F(x_1)]\,+\,[F(x_1)\,-\,F(x_0)] \, \end{matrix}$

The above quantity can be written as the following sum:

$F(b) - F(a) = \sum_{i=1}^n [F(x_i) - F(x_{i-1})] \qquad (1)$

Next we will employ the mean value theorem. Stated briefly,

Let F be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists some c in (a, b) such that

$F'(c) = \frac{F(b) - F(a)}{b - a}.$

It follows that

$F'(c)(b - a) = F(b) - F(a). \,$

The function F is differentiable on the interval [a, b]; therefore, it is also differentiable and continuous on each interval x_i-1. Therefore, according to the mean value theorem (above),

$F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}). \,$

Substituting the above into (1), we get

$F(b) - F(a) = \sum_{i=1}^n [F'(c_i)(x_i - x_{i-1})].$

The assumption implies $F'(c i) = f (c i).$ Also, $x i - x i - 1$ can be expressed as $Δ x$ of partition $i$ .

$F(b) - F(a) = \sum_{i=1}^n [f(c_i)(\Delta x_i)] \qquad (2)$

A converging sequence of Riemann sums. The numbers in the upper right are the areas of the grey rectangles. They converge to the integral of the function.

Notice that we are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the Mean Value Theorem, describes an approximation of the curve section it is drawn over. Also notice that $Δ x i$ does not need to be the same for any value of $i$ , or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with $n$ rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we will get closer and closer to the actual area of the curve.

By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann integral. That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity.

So, we take the limit on both sides of (2). This gives us

$\lim_{\| \Delta \| \to 0} F(b) - F(a) = \lim_{\| \Delta \| \to 0} \sum_{i=1}^n [f(c_i)(\Delta x_i)]\,.$

Neither F(b) nor F(a) is dependent on ||Δ||, so the limit on the left side remains F(b) - F(a).

$F(b) - F(a) = \lim_{\| \Delta \| \to 0} \sum_{i=1}^n [f(c_i)(\Delta x_i)]$

The expression on the right side of the equation defines an integral over f from a to b. Therefore, we obtain

$F(b) - F(a) = \int_{a}^{b} f(x)\,\mathrm dx$

which completes the proof.

We don't need to assume continuity of f on the whole interval. Part I of the theorem then says: if f is any Lebesgue integrable function on $[a, b]$ and $x 0$ is a number in $[a, b]$ such that $f$ is continuous at $x 0$ , then

$F(x) = \int_a^x f(t)\, \mathrm dt$

is differentiable for $x = x 0$ with $F'(x 0) = f (x 0)$ . We can relax the conditions on f still further and suppose that it is merely locally integrable. In that case, we can conclude that the function F is differentiable almost everywhere and F'(x)=f(x) almost everywhere. This is sometimes known as Lebesgue's differentiation theorem.

Part II of the theorem is true for any Lebesgue integrable function f which has an antiderivative F (not all integrable functions do, though).

The version of Taylor's theorem which expresses the error term as an integral can be seen as a generalization of the Fundamental Theorem.

There is a version of the theorem for complex functions: suppose U is an open set in C and f: U -> C is a function which has a holomorphic antiderivative F on U. Then for every curve γ : [a, b] -> U, the curve integral can be computed as

$\int_{\gamma} f(z) \,\mathrm dz = F(\gamma(b)) - F(\gamma(a)).$

The fundamental theorem can be generalized to curve and surface integrals in higher dimensions and on manifolds.

One of the most powerful statements in this direction is Stokes' theorem.

Differintegral

^ See, e.g., Marlow Anderson, Victor J. Katz, Robin J. Wilson, Sherlock Holmes in Babylon and Other Tales of Mathematical History, Mathematical Association of America, 2004, p. 114.

Larson, Ron, Bruce H. Edwards, David E. Heyd. Calculus of a single variable. 7th ed. Boston: Houghton Mifflin Company, 2002.
Leithold, L. (1996). The calculus 7 of a single variable. 6th ed. New York: HarperCollins College Publishers.
Malet, A, Studies on James Gregorie (1638-1675) (PhD Thesis, Princeton, 1989).
Stewart, J. (2003). Fundamental Theorem of Calculus. In Integrals. In Calculus: early transcendentals. Belmont, California: Thomson/Brooks/Cole.

Turnbull, H W (ed.), The James Gregory Tercentenary Memorial Volume (London, 1939)

Retrieved from "http://en.wikipedia.org../../../f/u/n/Fundamental_theorem_of_calculus.html"

Categories: Calculus | Mathematical theorems

Searching reference...

Reference

Sources

Fundamental theorem of calculus

From Wikipedia, the free encyclopedia

Contents

Top Matching Results

Sponsored Links
This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.

Search Results

Searching reference...

Reference

Sources

Fundamental theorem of calculus

From Wikipedia, the free encyclopedia

Contents

Top Matching Results

Sponsored Links This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.

Search Results

Sponsored Links
This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.